Time and Work problems $4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days. after this $1$ boy and $2$ girls are added and $\frac{1}{3}$ work is done in $3$ days. how many boys must be added to complete the remaining work in $1$ day?
My Attempt 
$4$ boys and $5$ girls can do $\frac{1}{2}$ work in $6$ days.
$5$ boys and $7$ girls can do $\frac{1}{3}$ work in $3$ days.
$5+x$ boys and $7$ girls can do $\frac{1}{6}$ work in $1$ day.
I could just reach upto here. can anyone help me further?
 A: You will need to figure out the amount of work done by one boy $b$ and one girl $g$. You have:
$$
\begin{array}{rcl}
(4b+5g)\times 6&=&\frac12,\\
(5b+7g)\times3&=&\frac13.
\end{array}
$$
Solving this for $b$ gives you $\frac{1}{108}$. The last of your items gives equation $((5+x)b+7g)\times 1 = \frac16$. This equation and the second give: $xb=\frac16-\frac19=\frac{1}{18}$ and, finally $x=\frac{108}{18}=6$.
A: Say
$b_w$: The amount of job can be done by a boy in one day
$g_w$: The amount of job can be done by a girl in one day
$(4b_w+5g_w)$ is the job done in one day, hence
All the job is $12(4b_w+5g_w)$ units
$1/3$ of the job can be done by 5 boys and 7 girls so all the job can also be expressed as $9(5b_w+7g_w)$.
$$9(5b_w+7g_w)=12(4b_w+5g_w)$$
$$b_w=g_w$$
(Boys and girls do the same job:)
Let's say 
$$b_w=g_w=x$$ to make things simpler. All the job is $108x$ units.
$1/6$ of the job remained at the end, which is $18x$. You want to finish this amount of job in one day; you have 7 girls who can do $7x$ , and you have 5 boys who can do $5x$, $18x-7x-5x=6x$ of job remains so you need 6 boys.
A: Let $b$  be the time (in days) a boy needs alone to complete the entire job and $g$ the time (in days) a girl needs alone to do the same. Then:
$$
4\frac{1}{b}+5\frac{1}{g}=\frac{\frac{1}{2}}{6}
$$
and also
$$
5\frac{1}{b}+7\frac{1}{g}=\frac{\frac{1}{3}}{3}
$$
which give $b=g=108$. Thus:
$$
(5+x)\frac{1}{108}+7\frac{1}{108}=\frac{1-\frac{1}{2}-\frac{1}{3}}{1}=\frac{\frac{1}{6}}{1}\Leftrightarrow 6(x+12)=108\Leftrightarrow x=6
$$
A: $1$ boy can do the work in $x$ days and $1$ girl do it in $y$ days.
In $1$ day, $1$ boy can do $1/x$ work.
In $6$ days, $4$ boys can do $24/x$ work.
In $6$ days, $5$ girls can do $30/y$ work.
$24/x + 30/y = 1/2$
One boy and two girls are added
$15/x + 21/y = 1/3$
$\Rightarrow x=y=108$
Boys and girls are equal.
$1/6$th of the work is left.
In $108$ days, $1$ work is completed by $1$ boy.
In $1$ day, $1$ work is completed by $108$ boys.
In $1$ day $1/6$ work is completed by $108/6 = 18$ boys.
$5$ boys and $7$ girls = $12$
Number of boys to be added to complete the remaining work in $1$ day 
$= 18 - 12 = 6$
